I was solving this question

If $\dfrac x3 + \dfrac 3x = 1$ then find the value of $x^3$.

I solved it as. Cube both sides and substitute $x^3$ with $t$, $$ \dfrac{t}{27} + \dfrac{27}{t}=-2$$ $$\implies t^2 + {27}^2 + 2\cdot 27 \cdot t = 0$$ $$\implies (t+27)^2=0$$ $$\implies t=x^3=-27$$

Someone over facebook put another solution as,

$$\dfrac x3 + \dfrac 3x = 1$$ $$\implies x^2-3x+9=0$$

Multiplying both the sides with $(x+3)$,

$$(x+3)(x^2-3x+9)=0$$ $$\implies x^3+27=0$$

Hence $x^3=-27$.

My question is, is it correct to multiply both sides with $0$? ( $x+3$ is $0$ here because $x^3=-27 \implies x=-3$ is one possible solution and other two being complex numbers.)

  • $\begingroup$ $x = -3$ is not a solution to the original equation. $\endgroup$ Nov 5, 2015 at 16:30
  • $\begingroup$ I didn't notice that. Suppose we are given that $x^2-4x+4=0$ then can we multiply both sides with $x+2$ and conclude $x^3=-8$? $\endgroup$
    – user103816
    Nov 5, 2015 at 16:34
  • $\begingroup$ Take a simpler example. Let $f(x) = x$ and we want to find the solutions to $x=0$ (which is just $0$ as you know; i.e. the origin of the straight line). Now multiply both sides by $(x+1)$ to get $(x+1)*x = 0$. But is $x = -1$ a solution to the original function? $\endgroup$ Nov 5, 2015 at 16:35
  • $\begingroup$ $(x+2)(x^2-4x+4)=0$ does not give $x^3=-8$. $\endgroup$
    – kmitov
    Nov 5, 2015 at 16:42
  • $\begingroup$ Ok. I mistook the formula $(a+b)(a^2+b^2-ab)$ $\endgroup$
    – user103816
    Nov 5, 2015 at 16:45

1 Answer 1


Yes there is nothing as such wrong with this process since it really helps in identifying the roots.

BUT one thing to keep in mind is that when you multiply the expression with $(x+3)$, you are introducing extraneous roots. That is, you had a quadratic equation with 2 roots and you have now made it cubic with 3 roots. So after solving, you must carefully eradicate this extraneous root you had introduced.

In the given problem, $x=-3$ is the extraneous root.

However you could have approached the problem using the discriminant.Then you get the roots easily.

$$x^2-3x+9=0 \Rightarrow x=\frac{3 \pm \sqrt{9-36}}{2}=\frac{3 \pm 3\sqrt{3}i}{2}$$

  • $\begingroup$ Now, if you rise $\frac{3 \pm i\sqrt{3}}{2}$ to the power 3, you will get $x^3=-27$. $\endgroup$
    – kmitov
    Nov 5, 2015 at 16:38
  • $\begingroup$ @kmitov What do you mean?....:-) $\endgroup$ Nov 5, 2015 at 16:39
  • $\begingroup$ PErhaps he meant does cubing $ x=\frac{3 \pm \sqrt{9-36}}{2}=\frac{3 \pm 3\sqrt{3}i}{2}$ give $-27$? $\endgroup$
    – user103816
    Nov 5, 2015 at 16:40
  • 1
    $\begingroup$ But the question is "what is the value of $x^3$?" $\endgroup$
    – kmitov
    Nov 5, 2015 at 16:44
  • 1
    $\begingroup$ No, you can't. Because $(x+2)(x^2-4x+4)=(x+2)(x-2)^2=(x^2-4)(x-2)=x^3-2x^2-4x+8 \not = x^3+8$. $\endgroup$ Nov 5, 2015 at 16:46

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